Class 12 Mathematics

OnlyGyaani • Class 12 Learning

Class 12 – Mathematics

Master 30 advanced topics covering advanced functions, inverse trigonometry, matrices, determinants, differential and integral calculus, differential equations, vectors, three-dimensional geometry, linear programming and probability. Every topic now includes detailed theory, key formulas, a fully worked example, common mistakes and practice questions.

1. Advanced Relations2. One-One & Onto3. Composition4. Inverse Functions5. Inverse Trigonometry6. Matrices7. Matrix Operations8. Inverse Matrix9. Determinants10. Linear Systems11. Continuity12. Differentiability13. Advanced Derivatives14. Tangents & Normals15. Increasing & Decreasing16. Maxima & Minima17. Indefinite Integrals18. Definite Integrals19. Integration Methods20. Area by Integration21. Differential Equations22. Vector Algebra23. Dot Product24. Cross Product25. Lines in 3D26. Planes in 3D27. Angles & Distances28. Linear Programming29. Conditional Probability30. Bayes & DistributionsFormula Sheet & Exam Tips

📘 How to use this page

Mathematics is learned by doing. Read the theory, memorise the formula box, then cover the solution of each worked example and try it yourself first. Study the “Common mistake” boxes, which show where marks are most often lost, and finish every topic by solving the practice questions on paper. Revise the formula sheet at the end before tests.

🔗 Advanced Relations, Functions and Inverse Trigonometry

1

Advanced Relations and Equivalence Relations

A relation R from a set A to a set B is a subset of A × B. A relation on A is a subset of A × A. Special cases are the empty relation (no pair related) and the universal relation (every pair related). Three properties classify relations on a set A:

  • Reflexive: (a, a) ∈ R for every a ∈ A.
  • Symmetric: (a, b) ∈ R ⇒ (b, a) ∈ R.
  • Transitive: (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R.

A relation that is reflexive, symmetric and transitive is an equivalence relation. It splits the set into disjoint equivalence classes: every element belongs to exactly one class, and elements in the same class are all related to each other. Congruence of triangles, similarity, equality, and “same remainder on division by n” are standard equivalence relations. In contrast, “is less than”, “is a divisor of” and “is perpendicular to” fail at least one of the three properties.

Counting facts (|A| = n): total relations = 2n²  |  reflexive relations = 2n²−n  |  symmetric relations = 2n(n+1)/2
Worked example: On the set of integers define aRb if (a − b) is divisible by 3. Show R is an equivalence relation.
Reflexive: a − a = 0 is divisible by 3. Symmetric: if 3 | (a − b) then 3 | (b − a) = −(a − b). Transitive: if 3 | (a − b) and 3 | (b − c), then 3 | [(a − b) + (b − c)] = (a − c). The equivalence classes are [0], [1], [2], consisting of integers leaving remainders 0, 1, 2 on division by 3.
Common mistake: Believing that symmetry plus transitivity automatically give reflexivity. For A = {1, 2, 3} and R = {(1,1), (1,2), (2,1), (2,2)}, R is symmetric and transitive, but (3, 3) ∉ R, so it is not reflexive.
Practice:
  1. Test whether “has the same remainder when divided by 3” is an equivalence relation on integers.
  2. Show that “is a divisor of” on natural numbers is reflexive and transitive but not symmetric.
  3. How many reflexive relations exist on a set with 3 elements? (answer: 2⁶ = 64)
Remember: An equivalence relation must satisfy all three conditions: reflexive, symmetric and transitive.
2

One-One, Onto and Bijective Functions

A function f: A → B assigns to every element of A exactly one element of B. A is the domain, B the codomain, and the set of actual outputs is the range. A function is one-one (injective) if different inputs give different outputs: f(x₁) = f(x₂) ⇒ x₁ = x₂. It is onto (surjective) if every element of the codomain is an output, i.e. range = codomain. A function that is both is bijective.

The graphical horizontal line test shows injectivity (no horizontal line meets the graph twice), while surjectivity depends on whether the range fills the stated codomain. Notice that the same formula can be bijective or not depending on the sets chosen: f(x) = x² is neither one-one nor onto from R to R, but it is a bijection from [0, ∞) to [0, ∞).

Counting facts (|A| = m, |B| = n): number of functions = nm  |  one-one functions = n!/(n − m)! (n ≥ m)  |  bijections = n! (when m = n)
Worked example: Show that f: R → R, f(x) = 2x + 1 is bijective.
One-one: 2x₁ + 1 = 2x₂ + 1 ⇒ x₁ = x₂. Onto: for any y ∈ R, take x = (y − 1)/2 ∈ R; then f(x) = y. Hence bijective. By contrast, f(x) = x² is not one-one since f(2) = f(−2) = 4, and not onto since negative numbers are never reached.
Common mistake: Judging “onto” without looking at the codomain. Always compare the range with the stated codomain.
Practice:
  1. Determine whether f(x) = x² from R to R is one-one or onto.
  2. How many one-one functions exist from {1, 2, 3} to {a, b, c, d, e}? (answer: 5 × 4 × 3 = 60)
  3. Check whether f(x) = x³ and f(x) = |x| are bijective on R.
Remember: A function has an inverse function on its stated sets exactly when it is bijective.
3

Composition of Functions

If f: A → B and g: B → C, the composite g∘f: A → C is defined by (g∘f)(x) = g(f(x)). The output of the inner function must lie in the domain of the outer function, so domains deserve care. Composition is associative, h∘(g∘f) = (h∘g)∘f, but generally not commutative: f∘g ≠ g∘f.

Useful structural facts: the composition of two one-one functions is one-one; the composition of two onto functions is onto; if g∘f is one-one then f must be one-one; and if g∘f is onto then g must be onto. The identity function I(x) = x behaves like the number 1: f∘I = I∘f = f.

Key facts: (f∘g)(x) = f(g(x))  |  domain of f∘g = {x in domain of g : g(x) in domain of f}  |  h∘(g∘f) = (h∘g)∘f
Worked example: Let f(x) = x + 2 and g(x) = 3x. Find f∘g and g∘f. Then, for f(x) = √x and g(x) = x − 4, find f∘g with its domain.
(f∘g)(x) = f(3x) = 3x + 2, whereas (g∘f)(x) = g(x + 2) = 3x + 6, so f∘g ≠ g∘f. For the second pair, (f∘g)(x) = √(x − 4), which requires x − 4 ≥ 0, so the domain is x ≥ 4.
Common mistake: Reading f∘g as “apply f first”. The function written on the right acts first.
Practice:
  1. Find g∘f for f(x) = x + 2 and g(x) = 3x, and compare it with f∘g.
  2. If f(x) = x² and g(x) = sin x, find f∘g and g∘f.
  3. If f(x) = 2x + 3 and g(x) = (x − 3)/2, show that f∘g = g∘f = identity.
Remember: Function composition is generally not commutative.
4

Inverse Functions

If f: A → B is bijective, there is a unique function f⁻¹: B → A such that f⁻¹(f(x)) = x for all x in A and f(f⁻¹(y)) = y for all y in B. To find it: write y = f(x), solve for x in terms of y, then rename y as x. The graph of f⁻¹ is the mirror image of the graph of f in the line y = x.

Properties: (f⁻¹)⁻¹ = f, and for bijective f and g, (g∘f)⁻¹ = f⁻¹∘g⁻¹ (the order reverses, like taking off shoes and socks). The notation f⁻¹(x) is not the reciprocal 1/f(x); the reciprocal is written [f(x)]⁻¹.

Method: y = f(x) ⟹ x = f⁻¹(y)  |  f∘f⁻¹ = f⁻¹∘f = I  |  (g∘f)⁻¹ = f⁻¹∘g⁻¹
Worked example: Find the inverse of f(x) = (x − 2)/(x + 3) (x ≠ −3).
Let y = (x − 2)/(x + 3). Then y(x + 3) = x − 2, so xy + 3y = x − 2, giving x(y − 1) = −2 − 3y and x = (3y + 2)/(1 − y). Therefore f⁻¹(x) = (3x + 2)/(1 − x). Simpler case: for f(x) = 3x − 6, f⁻¹(x) = (x + 6)/3.
Common mistake: Looking for an inverse of a function that is not bijective, such as f(x) = x² on all of R. Restrict the domain first.
Practice:
  1. Find the inverse of f(x) = 5x + 4 and verify by composition.
  2. Show that f(x) = x³ + 1 is invertible and find f⁻¹.
  3. If f(x) = 2x and g(x) = x + 3, find (g∘f)⁻¹.
Remember: f⁻¹(x) does not mean 1/f(x).
5

Inverse Trigonometric Functions

Trigonometric functions are many-one over R, so to define inverses their domains are restricted to intervals where they are bijective. The resulting outputs are principal values.

FunctionDomainPrincipal value range
sin⁻¹ x[−1, 1][−π/2, π/2]
cos⁻¹ x[−1, 1][0, π]
tan⁻¹ xR(−π/2, π/2)
cot⁻¹ xR(0, π)
sec⁻¹ x|x| ≥ 1[0, π] − {π/2}
cosec⁻¹ x|x| ≥ 1[−π/2, π/2] − {0}
Key identities:
sin⁻¹(−x) = −sin⁻¹x  |  cos⁻¹(−x) = π − cos⁻¹x  |  tan⁻¹(−x) = −tan⁻¹x
sin⁻¹x + cos⁻¹x = π/2  |  tan⁻¹x + cot⁻¹x = π/2
tan⁻¹x + tan⁻¹y = tan⁻¹[(x + y)/(1 − xy)] if xy < 1
2tan⁻¹x = sin⁻¹[2x/(1 + x²)] (|x| ≤ 1) = cos⁻¹[(1 − x²)/(1 + x²)] (x ≥ 0) = tan⁻¹[2x/(1 − x²)] (|x| < 1)
sin(sin⁻¹x) = x for x ∈ [−1, 1], but sin⁻¹(sin x) = x only for x ∈ [−π/2, π/2]
Worked example: Evaluate sin⁻¹(sin 2π/3) and tan⁻¹(1/2) + tan⁻¹(1/3).
2π/3 lies outside [−π/2, π/2], and sin(2π/3) = sin(π/3), so sin⁻¹(sin 2π/3) = π/3. Next, xy = 1/6 < 1, so the sum = tan⁻¹[(1/2 + 1/3)/(1 − 1/6)] = tan⁻¹[(5/6)/(5/6)] = tan⁻¹(1) = π/4. Also cos⁻¹(−1/2) = π − π/3 = 2π/3.
Common mistake: Writing sin⁻¹(sin x) = x for every x. The answer must lie in the principal range.
Practice:
  1. Evaluate cos⁻¹(0), tan⁻¹(1) and sin⁻¹(−1).
  2. Prove that sin⁻¹(3/5) + sin⁻¹(4/5) = π/2.
  3. Simplify tan⁻¹[(cos x)/(1 − sin x)], 0 < x < π/2 (answer: π/4 + x/2).
Remember: Principal-value ranges are essential when working with inverse trigonometric functions.

🔢 Matrices and Determinants

6

Matrices and Special Matrix Types

A matrix is a rectangular array of numbers arranged in m rows and n columns; its order is m × n and it has mn entries aij. Two matrices are equal when they have the same order and every corresponding entry is equal. Types include row matrix (1 × n), column matrix (m × 1), square matrix (m = n), diagonal (all off-diagonal entries zero), scalar (diagonal with equal entries), identity I (scalar with ones), zero matrix, and upper/lower triangular matrices.

For a square matrix A, the transpose A′ swaps rows and columns. A is symmetric if A′ = A and skew-symmetric if A′ = −A; a skew-symmetric matrix has zero diagonal entries. Every square matrix can be written uniquely as the sum of a symmetric and a skew-symmetric matrix.

Key facts: A = ½(A + A′) + ½(A − A′)  |  symmetric part ½(A + A′), skew part ½(A − A′)  |  a matrix of order m × n has mn entries
Worked example: Express A = [[1, 2], [3, 4]] as the sum of symmetric and skew-symmetric matrices.
A′ = [[1, 3], [2, 4]]. Symmetric part P = ½(A + A′) = [[1, 2.5], [2.5, 4]]. Skew part Q = ½(A − A′) = [[0, −0.5], [0.5, 0]]. Adding P + Q returns [[1, 2], [3, 4]] = A.
Common mistake: Comparing matrices of different orders and calling them equal because their entries look similar.
Practice:
  1. Classify several matrices by order and special type.
  2. How many different matrices of order 2 × 3 can be formed using entries 0 or 1? (answer: 2⁶ = 64)
  3. Show that A + A′ is symmetric and A − A′ is skew-symmetric for any square A.
Remember: Two matrices are equal only when they have the same order and corresponding entries are equal.
7

Matrix Operations and Multiplication

Matrices of the same order are added or subtracted entry by entry, and multiplied by a scalar entry by entry. Multiplication AB is defined only when the number of columns of A equals the number of rows of B: an m × n matrix times an n × p matrix gives an m × p matrix, where entry (i, j) is the sum of products of row i of A with column j of B.

Multiplication is associative and distributes over addition, but it behaves unlike ordinary numbers: AB ≠ BA in general, AB = O does not imply A = O or B = O, and AB = AC does not imply B = C. The transpose obeys (AB)′ = B′A′ with reversed order. Identity matrices satisfy AI = IA = A.

Key facts: (m × n)(n × p) = (m × p)  |  A(BC) = (AB)C  |  A(B + C) = AB + AC  |  (AB)′ = B′A′
(A + B)² = A² + AB + BA + B² (not A² + 2AB + B² unless AB = BA)
Worked example: For A = [[1, 2], [3, 4]] and B = [[0, 1], [1, 0]], compute AB and BA.
AB: row 1 = [1·0 + 2·1, 1·1 + 2·0] = [2, 1]; row 2 = [3·0 + 4·1, 3·1 + 4·0] = [4, 3]. So AB = [[2, 1], [4, 3]]. BA = [[3, 4], [1, 2]] (B swaps the rows of A). Clearly AB ≠ BA. Also [[1,0],[0,0]] × [[0,0],[0,1]] = O although neither factor is zero.
Common mistake: Expanding (A + B)² as A² + 2AB + B². Matrix multiplication is not commutative.
Practice:
  1. Determine whether AB and BA are defined for matrices of orders 2 × 3 and 3 × 2, and find their orders.
  2. If A = [[1, −1], [2, 3]], find A² − 4A + 5I.
  3. Show that (AB)′ = B′A′ for two 2 × 2 matrices of your choice.
Remember: Matrix multiplication is generally not commutative: AB need not equal BA.
8

Transpose and Inverse of a Matrix

The transpose satisfies (A′)′ = A, (A + B)′ = A′ + B′, (kA)′ = kA′ and (AB)′ = B′A′. A square matrix A of order n is invertible if there is a matrix B with AB = BA = I; then B = A⁻¹. This is possible exactly when A is non-singular (|A| ≠ 0). The inverse, when it exists, is unique.

The standard formula uses the adjoint: adj A is the transpose of the cofactor matrix, and A⁻¹ = adj A / |A|. For a 2 × 2 matrix [[a, b], [c, d]], swap a and d, change signs of b and c, and divide by ad − bc. Other facts: (AB)⁻¹ = B⁻¹A⁻¹, (A′)⁻¹ = (A⁻¹)′, A(adj A) = |A|I, and |adj A| = |A|n−1. Inverses can also be found by elementary row operations on [A | I].

Key formulas: A⁻¹ = adj A / |A|  |  for 2 × 2: A⁻¹ = (1/(ad − bc)) [[d, −b], [−c, a]]  |  (AB)⁻¹ = B⁻¹A⁻¹
Worked example: Find the inverse of A = [[2, 1], [5, 3]] and verify.
|A| = 2·3 − 1·5 = 1. A⁻¹ = [[3, −1], [−5, 2]]. Check: A·A⁻¹ = [[2·3 + 1·(−5), 2·(−1) + 1·2], [5·3 + 3·(−5), 5·(−1) + 3·2]] = [[1, 0], [0, 1]] = I.
Common mistake: Forgetting to transpose the cofactor matrix when forming the adjoint, or trying to invert a matrix with zero determinant.
Practice:
  1. Verify an inverse by multiplying the original matrix by the proposed inverse.
  2. Find the inverse of [[1, 2], [3, 4]] (answer: [[−2, 1], [1.5, −0.5]]).
  3. If A is 3 × 3 with |A| = 4, find |adj A| and |2A|. (answers: 16 and 32)
Remember: A matrix with determinant zero has no ordinary inverse.
9

Determinants and Their Properties

Each square matrix has a determinant, a single number. For 2 × 2, |A| = ad − bc. For 3 × 3, expand along any row or column using minors Mij and cofactors Cij = (−1)i+jMij, with signs following the checkerboard pattern + − + / − + − / + − +. Expanding along a row or column with many zeros saves time.

Properties: |A′| = |A|; interchanging two rows (or columns) changes the sign; two identical or proportional rows give 0; multiplying one row by k multiplies the determinant by k; adding a multiple of one row to another leaves it unchanged; |kA| = kⁿ|A| for an n × n matrix; and |AB| = |A||B|. A determinant gives geometric information too: the area of a triangle with vertices (x₁, y₁), (x₂, y₂), (x₃, y₃) is ½ |det [[x₁, y₁, 1], [x₂, y₂, 1], [x₃, y₃, 1]]|, and three points are collinear when this determinant is zero.

Key formulas: |A| = Σ aijCij (along any row/column)  |  |AB| = |A||B|  |  |kA| = kⁿ|A|
Area of triangle = ½ |det [[x₁, y₁, 1], [x₂, y₂, 1], [x₃, y₃, 1]]|
Worked example: Evaluate |[[1, 2, 3], [0, 1, 4], [5, 6, 0]]| and the area of the triangle with vertices (0,0), (4,0), (0,3).
Expanding along row 1: 1(1·0 − 4·6) − 2(0·0 − 4·5) + 3(0·6 − 1·5) = −24 + 40 − 15 = 1. For the triangle, det [[0,0,1],[4,0,1],[0,3,1]] = 1·(4·3 − 0) = 12, so area = ½ × 12 = 6 square units.
Common mistake: Losing the sign pattern of cofactors, or using |kA| = k|A| for a 3 × 3 matrix (the correct factor is k³).
Practice:
  1. Evaluate the determinant of a 3 × 3 matrix by expansion along a convenient row.
  2. Show that [[1, a, a²], [1, b, b²], [1, c, c²]] has determinant (a − b)(b − c)(c − a) up to sign.
  3. Find k if the points (1, 2), (3, 4) and (k, 6) are collinear. (answer: k = 5)
Remember: Interchanging two rows or columns changes the sign of a determinant.
10

Solving Systems of Linear Equations with Matrices

A system of linear equations is written compactly as AX = B, where A is the coefficient matrix, X the column of unknowns and B the column of constants. If |A| ≠ 0, A is invertible and the system has a unique solution X = A⁻¹B (equivalently Cramer’s rule: x = Dx/D, y = Dy/D, z = Dz/D, where Dx replaces the x-column with B).

If |A| = 0, compute (adj A)B. If it is the zero matrix, the system is consistent with infinitely many solutions; if it is non-zero, the system is inconsistent (no solution). A system that has at least one solution is called consistent.

Key facts: |A| ≠ 0 ⇒ unique solution X = A⁻¹B  |  |A| = 0 and (adj A)B = O ⇒ infinitely many  |  |A| = 0 and (adj A)B ≠ O ⇒ none
Worked example: Solve x + y + z = 6, x − y + z = 2, 2x + y − z = 1.
D = |[[1,1,1],[1,−1,1],[2,1,−1]]| = 1(1 − 1) − 1(−1 − 2) + 1(1 + 2) = 0 + 3 + 3 = 6 ≠ 0, so the solution is unique. Subtracting equations 1 and 2 gives 2y = 4, y = 2; adding them gives x + z = 4; then equation 3 gives 2x − z = −1, hence x = 1, z = 3. Check in the first equation: 1 + 2 + 3 = 6.
Inconsistent case: x + y = 2 and 2x + 2y = 5 have |A| = 0 but contradictory constants, so no solution.
Common mistake: Applying Cramer’s rule when D = 0. The rule needs a non-zero determinant.
Practice:
  1. Convert a three-equation system into matrix form AX = B.
  2. Solve 2x + 3y = 7 and x − y = 1 using the matrix inverse (answer: x = 2, y = 1).
  3. Find k for which x + 2y = 3 and 2x + ky = 6 has infinitely many solutions (answer: k = 4).
Remember: Matrix methods reveal both the structure and solvability of linear systems.

📈 Differential Calculus and Applications

11

Continuity of Functions

A function f is continuous at x = c if limx→c f(x) exists and equals f(c); that is, left-hand limit = right-hand limit = f(c). It is continuous on an interval if continuous at every point of it. Intuitively, the graph can be drawn without lifting the pen.

Sums, differences, products, quotients (where the denominator is non-zero) and compositions of continuous functions are continuous. Polynomials, sine, cosine, exponential functions and absolute value are continuous everywhere on their domains; rational functions are continuous except where the denominator is zero. A discontinuity is removable if the limit exists but differs from (or the function lacks) the value there, and a jump discontinuity if the one-sided limits are different (the greatest integer function jumps at every integer).

Conditions at x = c: LHL = limx→c⁻ f(x)  |  RHL = limx→c⁺ f(x)  |  continuous ⟺ LHL = RHL = f(c)
Worked example: Find k so that f(x) = (x² − 4)/(x − 2) for x ≠ 2 and f(2) = k is continuous at 2. Then find k so that f(x) = kx + 1 (x ≤ 5), 3x − 5 (x > 5) is continuous at 5.
First: for x ≠ 2, f(x) = x + 2, so the limit is 4; therefore k = 4. Second: RHL = 3(5) − 5 = 10 and LHL = f(5) = 5k + 1; equating gives 5k + 1 = 10, so k = 9/5.
Common mistake: Checking only that the limit exists. The function value at the point must also equal that limit.
Practice:
  1. Test continuity of a piecewise function at its joining point.
  2. Show that f(x) = |x| is continuous at 0.
  3. Find the points of discontinuity of f(x) = 1/(x² − 3x + 2). (answer: x = 1 and x = 2)
Remember: Continuity at x = a requires lim f(x) = f(a).
12

Differentiability and Its Relationship with Continuity

The derivative of f at c is f′(c) = limh→0 [f(c + h) − f(c)]/h, provided the limit exists and is finite. It exists only if the left-hand derivative (LHD) and right-hand derivative (RHD) are equal. Geometrically, differentiability means the graph has a unique non-vertical tangent there. Corners (like |x| at 0), cusps and vertical tangents (like x1/3 at 0) violate it.

The central relationship is: differentiable ⇒ continuous, so a function that is not continuous at a point cannot be differentiable there. The converse is false: |x| is continuous at 0 yet not differentiable. For piecewise functions, check continuity first, then compare LHD and RHD at the joining point.

Definitions: f′(c) = limh→0 [f(c + h) − f(c)]/h  |  differentiable at c ⟺ LHD = RHD (finite)  |  differentiable ⇒ continuous
Worked example: Examine f(x) = |x| at x = 0, and f(x) = x² (x ≤ 1), 2x − 1 (x > 1) at x = 1.
For |x|: LHD = lim (−h − 0)/h as h→0⁻ = −1, RHD = +1; unequal, so not differentiable at 0. For the second function: f(1) = 1, RHL = 2(1) − 1 = 1, so f is continuous. LHD = 2(1) = 2 (derivative of x²) and RHD = 2, equal; so it is differentiable at 1 with f′(1) = 2.
Common mistake: Assuming a continuous function must be differentiable. Sharp corners break differentiability.
Practice:
  1. Examine left and right derivatives of |x| at zero.
  2. Show that f(x) = |x − 2| is continuous but not differentiable at x = 2.
  3. Is f(x) = x|x| differentiable at 0? (answer: yes, f′(0) = 0)
Remember: Differentiable ⇒ continuous, but continuous ⇏ differentiable.
13

Advanced Differentiation Techniques

Beyond basic rules, Class 12 uses the chain rule for composites, the product and quotient rules, implicit differentiation (differentiate both sides with respect to x and treat y as a function of x), logarithmic differentiation (for functions like xx or products with many factors), parametric differentiation and second-order derivatives. Rolle’s theorem and the mean value theorem link average and instantaneous rates of change.

Standard derivatives:
d/dx(xⁿ) = nxⁿ⁻¹  |  sin x → cos x  |  cos x → −sin x  |  tan x → sec²x  |  eˣ → eˣ  |  ln x → 1/x
sin⁻¹x → 1/√(1 − x²)  |  cos⁻¹x → −1/√(1 − x²)  |  tan⁻¹x → 1/(1 + x²)  |  cot⁻¹x → −1/(1 + x²)
Chain rule: dy/dx = (dy/du)(du/dx)  |  Parametric: dy/dx = (dy/dt)/(dx/dt)  |  Product: (uv)′ = u′v + uv′
Worked example: Differentiate (a) y = (3x² + 1)⁵, (b) y = xx, (c) x² + y² = 25, (d) x = a cos θ, y = a sin θ.
(a) dy/dx = 5(3x² + 1)⁴ · 6x = 30x(3x² + 1)⁴. (b) ln y = x ln x ⟹ y′/y = ln x + 1 ⟹ y′ = xx(1 + ln x). (c) 2x + 2y(dy/dx) = 0 ⟹ dy/dx = −x/y. (d) dy/dx = (a cos θ)/(−a sin θ) = −cot θ.
Common mistake: Forgetting the inner derivative in the chain rule, for instance differentiating sin(3x) as cos(3x) instead of 3cos(3x).
Practice:
  1. Differentiate y = (3x² + 1)⁵ using the chain rule.
  2. Find dy/dx if y = sin⁻¹(2x√(1 − x²)).
  3. If y = ex sin x, show that d²y/dx² − 2(dy/dx) + 2y = 0.
Remember: Identify the structure of a function before selecting a derivative rule.
14

Tangents and Normals to Curves

The derivative dy/dx at a point is the slope of the tangent to the curve there. If the point is (x₀, y₀) and m = f′(x₀), the tangent is y − y₀ = m(x − x₀) and the normal (perpendicular to the tangent at the same point) is y − y₀ = −(1/m)(x − x₀). If the tangent is horizontal (m = 0), the normal is the vertical line x = x₀; if the tangent is vertical, the normal is horizontal.

The tangent also gives approximations: for small Δx, f(x + Δx) ≈ f(x) + f′(x)Δx. This is used to estimate values such as square roots and small errors in measurements (Δy ≈ dy).

Key formulas: Tangent: y − y₀ = f′(x₀)(x − x₀)  |  Normal: y − y₀ = −[1/f′(x₀)](x − x₀)  |  Δy ≈ (dy/dx)Δx
Worked example: Find the tangent and normal to y = x² at x = 1, and approximate √25.3.
At x = 1, y = 1 and slope = 2x = 2. Tangent: y − 1 = 2(x − 1), i.e. y = 2x − 1. Normal: y − 1 = −½(x − 1), i.e. x + 2y = 3. For √25.3, take f(x) = √x at 25 with Δx = 0.3: f′ = 1/(2√x) = 1/10, so √25.3 ≈ 5 + 0.03 = 5.03.
Common mistake: Using the derivative formula before finding the y-coordinate, or using m instead of −1/m for the normal.
Practice:
  1. Find tangent and normal equations to y = x² at x = 1.
  2. Find the points on y = x³ − 3x where the tangent is parallel to the x-axis. (answer: x = ±1)
  3. Use differentials to approximate (27.1)1/3. (answer: ≈ 3.0037)
Remember: First find the point on the curve, then calculate the derivative there.
15

Increasing and Decreasing Functions

A function f is increasing on an interval if x₁ < x₂ implies f(x₁) ≤ f(x₂), and strictly increasing if the inequality is strict; the decreasing cases reverse the inequality. For a differentiable function, the sign of the derivative decides: f′(x) > 0 on an interval ⇒ strictly increasing; f′(x) < 0 ⇒ strictly decreasing; f′(x) ≥ 0 (with zeros only at isolated points) still gives increasing.

Procedure: find f′(x), solve f′(x) = 0 to locate critical points, mark these on a number line, test the sign of f′ in each sub-interval, and report open intervals. A function like x³ is increasing everywhere even though f′(0) = 0, because the derivative does not change sign.

Rule: f′(x) > 0 ⇒ f strictly increasing  |  f′(x) < 0 ⇒ f strictly decreasing  |  critical points: f′(x) = 0 or f′ undefined
Worked example: Find the intervals where f(x) = x³ − 3x increases or decreases.
f′(x) = 3x² − 3 = 3(x − 1)(x + 1). The critical points are x = −1 and x = 1. For x < −1: f′ > 0 (increasing). For −1 < x < 1: f′ < 0 (decreasing). For x > 1: f′ > 0 (increasing). So f increases on (−∞, −1) and (1, ∞) and decreases on (−1, 1).
Common mistake: Writing a single interval such as (−∞, ∞) for “increasing or decreasing” without testing sign changes of f′.
Practice:
  1. Determine increasing and decreasing intervals for f(x) = x³ − 3x.
  2. Show that f(x) = x³ + x is strictly increasing on R.
  3. Find the intervals on which f(x) = sin x is increasing in [0, 2π]. (answer: [0, π/2] and [3π/2, 2π])
Remember: Critical points help divide the domain into intervals for sign analysis.
16

Maxima, Minima and Optimisation

A point c is a local maximum if f(c) is the largest value in some neighbourhood of c, and a local minimum if it is the smallest. A necessary condition is f′(c) = 0 (or f′ undefined). The first derivative test classifies c: if f′ changes from positive to negative it is a local maximum; from negative to positive a local minimum; no sign change means a point of inflection. The second derivative test says: f′(c) = 0 and f″(c) < 0 ⇒ local maximum; f″(c) > 0 ⇒ local minimum; f″(c) = 0 ⇒ inconclusive.

On a closed interval [a, b] the absolute maximum and minimum occur at critical points inside the interval or at the endpoints, so evaluate f at all of them and compare. For word problems, express the quantity to be optimised in one variable using the constraint, find critical points, verify the type, and interpret the answer within the allowed domain.

Tests: f′(c) = 0, f″(c) < 0 ⇒ max  |  f′(c) = 0, f″(c) > 0 ⇒ min  |  Closed interval: compare f(a), f(b) and f at critical points
Worked example 1: A rectangle has perimeter 40. Find the maximum area.
With sides x and 20 − x, A = x(20 − x) = 20x − x². A′ = 20 − 2x = 0 ⟹ x = 10, and A″ = −2 < 0, so it is a maximum: A = 100 (a square).
Worked example 2: Absolute extrema of f(x) = x³ − 3x on [−2, 2]: critical points x = ±1. f(−2) = −2, f(−1) = 2, f(1) = −2, f(2) = 2; so the maximum is 2 and the minimum is −2.
Worked example 3: An open box is made by cutting squares of side x from the corners of an 18 cm square sheet. V = x(18 − 2x)²; V′ = (18 − 2x)(18 − 6x) = 0 ⟹ x = 3 (x = 9 gives zero volume). V = 3 × 12² = 432 cm³ (maximum).
Common mistake: Treating every stationary point as a maximum or minimum. Always classify it, and check endpoints on closed intervals.
Practice:
  1. Form an area function for a rectangle with fixed perimeter and determine its maximum.
  2. Find the local maximum and minimum values of f(x) = 2x³ − 9x² + 12x + 5.
  3. A closed cylinder has volume 1000 cm³. Find the radius that minimises surface area. (answer: r = (500/π)1/3, with h = 2r)
Remember: A stationary point is a candidate; use sign changes or another test to classify it.

∫ Integral Calculus and Differential Equations

17

Indefinite Integrals and Antiderivatives

Integration is the reverse of differentiation. If F′(x) = f(x), then F is an antiderivative of f and the indefinite integral is ∫f(x)dx = F(x) + C, where C is an arbitrary constant, because every member of the family F(x) + C has the same derivative. Integration is linear: ∫[af(x) + bg(x)]dx = a∫f + b∫g.

Standard integrals:
∫xⁿdx = xⁿ⁺¹/(n + 1) + C (n ≠ −1)  |  ∫dx/x = ln|x| + C  |  ∫eˣdx = eˣ + C  |  ∫aˣdx = aˣ/ln a + C
∫sin x dx = −cos x + C  |  ∫cos x dx = sin x + C  |  ∫sec²x dx = tan x + C  |  ∫cosec²x dx = −cot x + C
∫tan x dx = ln|sec x| + C  |  ∫cot x dx = ln|sin x| + C  |  ∫sec x dx = ln|sec x + tan x| + C  |  ∫cosec x dx = ln|cosec x − cot x| + C
∫dx/(x² + a²) = (1/a)tan⁻¹(x/a) + C  |  ∫dx/√(a² − x²) = sin⁻¹(x/a) + C
∫dx/(x² − a²) = (1/2a)ln|(x − a)/(x + a)| + C  |  ∫dx/√(x² ± a²) = ln|x + √(x² ± a²)| + C
Worked example: Evaluate ∫(3x² + 4)dx and ∫sin²x dx.
∫(3x² + 4)dx = x³ + 4x + C. For sin²x use sin²x = (1 − cos 2x)/2: ∫sin²x dx = x/2 − (sin 2x)/4 + C. Differentiating the answer returns the integrand, which is always a good check.
Common mistake: Omitting + C, or writing ∫dx/x = ln x without the absolute value.
Practice:
  1. Evaluate ∫(3x² + 4)dx.
  2. Evaluate ∫(2x + 3)/(x² + 3x + 5) dx (answer: ln|x² + 3x + 5| + C).
  3. Evaluate ∫1/(1 + cos 2x) dx (answer: ½ tan x + C).
Remember: Always include the constant of integration in an indefinite integral.
18

Definite Integrals and Fundamental Theorem

The definite integral ∫ab f(x)dx is the limit of Riemann sums and represents signed area between the curve and the x-axis. The second fundamental theorem of calculus says that if F is an antiderivative of f, then ∫ab f(x)dx = F(b) − F(a). The first fundamental theorem states that the derivative of A(x) = ∫ax f(t)dt is f(x).

Properties:
∫abf = −∫baf  |  ∫abf = ∫acf + ∫cbf
∫0af(x)dx = ∫0af(a − x)dx  |  ∫abf(x)dx = ∫abf(a + b − x)dx
∫−aaf = 2∫0af if f is even, 0 if f is odd
∫02af(x)dx = 2∫0af(x)dx if f(2a − x) = f(x), and 0 if f(2a − x) = −f(x)
Worked example: Evaluate ∫13 2x dx and I = ∫0π/2 sin x/(sin x + cos x) dx.
∫13 2x dx = [x²]13 = 9 − 1 = 8. For I, apply the property f(x) → f(π/2 − x): I = ∫ cos x/(cos x + sin x) dx. Adding the two forms gives 2I = ∫0π/2 1 dx = π/2, so I = π/4. Also ∫−11 x³dx = 0 because x³ is odd.
Common mistake: Substituting a new variable but forgetting to change the limits (or to change them back).
Practice:
  1. Evaluate ∫13 2x dx.
  2. Evaluate ∫0π x sin x dx (answer: π).
  3. Use the even/odd property to evaluate ∫−π/2π/2 sin⁵x cos x dx. (answer: 0)
Remember: A definite integral produces a number and does not require +C.
19

Methods of Integration

Substitution: if the integrand has the form f(g(x))g′(x), put t = g(x) so that ∫f(t)dt is simpler. Integration by parts comes from the product rule: ∫u v dx = u∫v dx − ∫[u′ ∫v dx]dx. Choose u using the order ILATE (Inverse trig, Logarithmic, Algebraic, Trigonometric, Exponential): the earlier-listed function is taken as u. Partial fractions decompose a rational function into simpler fractions that integrate separately; if the degree of the numerator is not smaller than the denominator, divide first.

Useful results:
∫eˣ[f(x) + f′(x)]dx = eˣf(x) + C  |  ∫√(a² − x²)dx = (x/2)√(a² − x²) + (a²/2)sin⁻¹(x/a) + C
∫√(x² + a²)dx = (x/2)√(x² + a²) + (a²/2)ln|x + √(x² + a²)| + C
Partial fractions: (px + q)/[(x − a)(x − b)] = A/(x − a) + B/(x − b)
Worked example: (a) ∫2x cos(x²)dx; (b) ∫x eˣdx; (c) ∫ln x dx; (d) ∫(x + 1)/[(x − 1)(x + 2)]dx.
(a) Put t = x²: ∫cos t dt = sin(x²) + C. (b) u = x, v = eˣ: x eˣ − ∫eˣdx = (x − 1)eˣ + C. (c) u = ln x, v = 1: x ln x − x + C. (d) x + 1 = A(x + 2) + B(x − 1) gives A = 2/3 and B = 1/3, so the integral is (2/3)ln|x − 1| + (1/3)ln|x + 2| + C. Also ∫eˣ(sin x + cos x)dx = eˣ sin x + C.
Common mistake: Choosing u and v the wrong way round, which makes the new integral harder. Use ILATE and test quickly.
Practice:
  1. Identify whether substitution or integration by parts is more natural for ∫x²eˣdx, ∫tan x dx and ∫x/(x² + 1)dx.
  2. Evaluate ∫x²eˣdx (answer: eˣ(x² − 2x + 2) + C).
  3. Evaluate ∫dx/(x² − 1) using partial fractions.
Remember: Simplifying the integrand before integrating can reveal an easier method.
20

Area Under Curves and Between Curves

For a continuous curve y = f(x) lying above the x-axis, the area bounded by the curve, the x-axis and the lines x = a and x = b is ∫ab f(x)dx. If part of the curve lies below the axis, split the integral at the x-intercepts and add absolute values. The area between two curves y = f(x) (upper) and y = g(x) (lower) over [a, b] is ∫ab[f(x) − g(x)]dx. First find intersection points by solving f(x) = g(x), then sketch the region. Sometimes it is easier to integrate with respect to y.

Standard areas: Circle x² + y² = a²: πa² (= 4∫0a√(a² − x²)dx)  |  Ellipse x²/a² + y²/b² = 1: πab
Region between y² = 4ax and its latus rectum x = a: (8/3)a²  |  Area = ∫(upper − lower)dx
Worked example: Find the area between y = x² and y = x, and the area enclosed by y² = 4x and y = x.
First: intersection x² = x ⟹ x = 0, 1; on [0, 1] the line is above the parabola: ∫01(x − x²)dx = ½ − ⅓ = 1/6. Second: y² = 4x and y = x meet where x² = 4x, so x = 0, 4; upper curve y = 2√x, lower y = x: ∫04(2√x − x)dx = (4/3)(4)3/2 − 8 = 32/3 − 8 = 8/3.
Common mistake: Subtracting in the wrong order, which gives a negative area. Area must be non-negative.
Practice:
  1. Sketch two curves before setting up the integral for their enclosed area.
  2. Find the area of the region bounded by y = sin x, the x-axis, x = 0 and x = 2π. (answer: 4)
  3. Find the area of the circle x² + y² = 16 using integration. (answer: 16π)
Remember: Geometric area is non-negative even though definite integrals can be signed.
21

Differential Equations and Variable Separation

A differential equation involves an unknown function and its derivatives. Its order is the order of the highest derivative; its degree is the power of that highest derivative once the equation is a polynomial in derivatives (not defined if derivatives appear inside trigonometric or exponential functions). A general solution contains arbitrary constants equal in number to the order; a particular solution uses initial conditions to fix them. A family of curves with n arbitrary constants is described by an nth-order differential equation formed by eliminating the constants.

Three solution methods are required. Variable separable: dy/dx = f(x)g(y) ⟹ ∫dy/g(y) = ∫f(x)dx. Homogeneous: dy/dx = F(y/x); substitute y = vx. Linear first-order: dy/dx + Py = Q, with integrating factor IF = e∫P dx, solution y·IF = ∫Q·IF dx + C (and similarly dx/dy + P₁x = Q₁).

Key results: IF = e∫P dx  |  y(IF) = ∫Q(IF)dx + C  |  dy/dx = ky ⟹ y = Cekx (growth for k > 0, decay for k < 0)
Worked example: (a) Solve dy/dx = xy with y(0) = 2. (b) Solve dy/dx + y = e⁻ˣ. (c) State order and degree of (d²y/dx²)³ + (dy/dx)² + y = 0.
(a) Separate: dy/y = x dx ⟹ ln|y| = x²/2 + C ⟹ y = Aex²/2; y(0) = 2 gives A = 2. (b) P = 1, IF = eˣ; (y eˣ)′ = 1 ⟹ y eˣ = x + C ⟹ y = (x + C)e⁻ˣ. (c) Highest derivative is second order, with power 3, so order 2 and degree 3.
Common mistake: Losing the constant of integration before applying the initial condition, or forgetting the absolute value in ln|y|.
Practice:
  1. Identify the order and degree of several differential equations.
  2. Solve dy/dx = (1 + y²)/(1 + x²) (answer: tan⁻¹y = tan⁻¹x + C).
  3. A population grows at a rate proportional to its size and doubles in 10 years. Write the differential equation and solution.
Remember: Initial conditions determine constants in a general solution.

🧭 Vectors and Three-Dimensional Geometry

22

Vector Algebra and Components

A vector has both magnitude and direction. The position vector of the point (x, y, z) is r = xî + yĵ + zk̂, with magnitude |r| = √(x² + y² + z²). The vector from P to Q is PQ = OQ − OP. A unit vector in the direction of a is â = a/|a|. Vectors add by the triangle or parallelogram law, and component-wise in coordinates. Two vectors are collinear (parallel) if one is a scalar multiple of the other.

The direction cosines l, m, n of a vector are the cosines of the angles α, β, γ it makes with the axes; l = x/|r|, and l² + m² + n² = 1. Direction ratios are any numbers proportional to them. The section formula locates the point dividing the join of A and B in the ratio m : n internally at (m·B + n·A)/(m + n), and the midpoint is (A + B)/2.

Key formulas: |a| = √(a₁² + a₂² + a₃²)  |  â = a/|a|  |  l = cos α = a₁/|a|  |  l² + m² + n² = 1  |  Internal division: (m·b + n·a)/(m + n)
Worked example: For a = 2î − ĵ + 2k̂ find |a|, â and its direction cosines. Then find the point dividing A(1, 2, 3) and B(7, 8, 9) in the ratio 1 : 2.
|a| = √(4 + 1 + 4) = 3, so â = (2/3, −1/3, 2/3), and these are also the direction cosines. Internal division: P = (1·B + 2·A)/3 = ((7 + 2)/3, (8 + 4)/3, (9 + 6)/3) = (3, 4, 5).
Common mistake: Adding magnitudes of vectors instead of adding vectors. |a + b| ≤ |a| + |b|, with equality only for parallel vectors in the same direction.
Practice:
  1. Find the magnitude and unit vector corresponding to (2, −1, 2).
  2. Show that the points A(1, 2, 3), B(3, 4, 5), C(5, 6, 7) are collinear.
  3. Find the direction cosines of the line joining (1, 0, 0) and (0, 1, 1).
Remember: A unit vector has magnitude 1 and indicates direction.
23

Dot Product and Projection

The scalar (dot) product of two vectors is a·b = |a||b|cos θ, where θ is the angle between them (0 ≤ θ ≤ π). In components, a·b = a₁b₁ + a₂b₂ + a₃b₃. It is commutative and distributive, with a·a = |a|², and î·î = ĵ·ĵ = k̂·k̂ = 1 while î·ĵ = ĵ·k̂ = k̂·î = 0. The product is zero for non-zero vectors exactly when they are perpendicular.

The scalar projection of a on b is (a·b)/|b|; the vector projection is [(a·b)/|b|²]b. Physically, the work done by a force F over displacement d is W = F·d.

Key formulas: cos θ = (a·b)/(|a||b|)  |  projection of a on b = (a·b)/|b|  |  |a + b|² = |a|² + |b|² + 2a·b  |  |a·b| ≤ |a||b|
Worked example: Find the angle between a = (1, 2, 2) and b = (2, −1, 2) and the projection of a on b. Also test whether (2, 3, −1) and (1, −1, −1) are perpendicular.
a·b = 2 − 2 + 4 = 4; |a| = 3, |b| = 3; cos θ = 4/9, so θ = cos⁻¹(4/9). Projection of a on b = 4/3. For the second pair: 2 − 3 + 1 = 0, so they are perpendicular.
Common mistake: Treating a·b as a vector. The dot product is a scalar.
Practice:
  1. Find the angle between two given vectors using their dot product.
  2. Find λ so that (2î + λĵ + k̂) ⊥ (î − 2ĵ + 3k̂). (answer: λ = 5/2)
  3. A force F = (3, 4, 5) moves a particle from (1, 1, 1) to (2, 3, 4). Find the work done. (answer: F·d = 3 + 8 + 15 = 26 units)
Remember: The dot product produces a scalar, not a vector.
24

Cross Product and Geometric Applications

The vector (cross) product is a × b = |a||b| sin θ n̂, where n̂ is a unit vector perpendicular to both, in the direction given by the right-hand rule. In components it is the determinant with rows î ĵ k̂ / a₁ a₂ a₃ / b₁ b₂ b₃. It is anti-commutative (a × b = −b × a), distributive, and a × a = 0; a × b = 0 for non-zero vectors means they are parallel.

Geometric uses: the area of a parallelogram with adjacent sides a and b is |a × b|; the area of a triangle is ½|a × b|. The scalar triple product [a b c] = a·(b × c) equals the determinant of the three vectors, and its absolute value is the volume of the parallelepiped. Three vectors are coplanar when it is zero. Lagrange’s identity: |a × b|² = |a|²|b|² − (a·b)².

Key formulas: |a × b| = |a||b| sin θ  |  Parallelogram area = |a × b|  |  Triangle area = ½|a × b|  |  [a b c] = det[a; b; c]  |  Volume = |[a b c]|
Worked example: For a = (1, 2, 3) and b = (2, −1, 1) find a × b and the area of the triangle with these sides.
î: 2·1 − 3·(−1) = 5; ĵ: 3·2 − 1·1 = 5; k̂: 1·(−1) − 2·2 = −5. So a × b = (5, 5, −5), |a × b| = 5√3, and the triangle area = (5√3)/2. Check: a·(a × b) = 5 + 10 − 15 = 0, confirming perpendicularity.
Common mistake: Reversing the order and ignoring the sign change. Also, do not use the cross product in a plane problem expecting a scalar.
Practice:
  1. Use a cross product to test whether two vectors are parallel.
  2. Find the area of the triangle with vertices A(1, 1, 1), B(1, 2, 3), C(2, 3, 1).
  3. Show that (1, 2, 3), (2, 1, 0), (3, 3, 3) are coplanar (the scalar triple product is zero).
Remember: Reversing cross-product order reverses direction: a × b = −(b × a).
25

Equations of Lines in Three Dimensions

A line is fixed by a point and a direction. Vector form: r = a + λb, where a is the position vector of a point on the line and b a direction vector. Through two points A and B: r = a + λ(b − a). Cartesian (symmetric) form: (x − x₁)/l = (y − y₁)/m = (z − z₁)/n, with direction ratios (l, m, n); the parameter λ gives the parametric form x = x₁ + λl, etc.

Angle between lines with direction vectors b₁ and b₂: cos θ = |b₁·b₂|/(|b₁||b₂|). Lines are parallel if b₁ ∥ b₂ and perpendicular if b₁·b₂ = 0. Skew lines are non-parallel and non-intersecting; their shortest distance is |(a₂ − a₁)·(b₁ × b₂)|/|b₁ × b₂|. For parallel lines, the distance is |(a₂ − a₁) × b|/|b|.

Key formulas: Skew-line distance d = |(a₂ − a₁)·(b₁ × b₂)| / |b₁ × b₂|  |  Parallel-line distance d = |(a₂ − a₁) × b| / |b|
Worked example 1: Find the line through A(1, 2, 3) and B(3, 5, 9). Direction = (2, 3, 6), so (x − 1)/2 = (y − 2)/3 = (z − 3)/6; |direction| = 7, so direction cosines are (2/7, 3/7, 6/7).
Worked example 2: Find the shortest distance between r = (1, 1, 0) + λ(2, −1, 1) and r = (2, 1, −1) + μ(3, −5, 2). b₁ × b₂ = (3, −1, −7), with magnitude √59; a₂ − a₁ = (1, 0, −1); dot product = 3 + 7 = 10. Distance = 10/√59.
Common mistake: Confusing direction ratios with coordinates of a point on the line, or dropping the absolute value in the angle formula.
Practice:
  1. Form the vector equation of a line through two given points.
  2. Find the angle between the lines (x − 1)/1 = (y − 2)/2 = (z − 3)/2 and (x)/2 = (y)/(−1) = (z)/2.
  3. Convert r = (1, 2, 3) + λ(2, 3, 4) into Cartesian form.
Remember: A point fixes location; a direction vector fixes orientation.
26

Equations of Planes in Three Dimensions

A plane is fixed by a point and a normal vector n. Vector form: r·n = d (or (r − a)·n = 0). Cartesian form: ax + by + cz = d, where (a, b, c) is a normal. Other forms: through the point (x₁, y₁, z₁) with normal (a, b, c): a(x − x₁) + b(y − y₁) + c(z − z₁) = 0; intercept form x/p + y/q + z/r = 1; through three non-collinear points, using the determinant of coordinate differences or the normal (AB × AC); and through the intersection of two planes P₁ + λP₂ = 0.

Key formulas: Plane: ax + by + cz = d, normal (a, b, c)  |  Intercept form: x/p + y/q + z/r = 1  |  Through intersection: (a₁x + b₁y + c₁z − d₁) + λ(a₂x + b₂y + c₂z − d₂) = 0
Worked example: Find the plane through (1, 2, 3) with normal (2, −1, 3), and the plane through (1, 0, 0), (0, 1, 0), (0, 0, 1).
2(x − 1) − (y − 2) + 3(z − 3) = 0 gives 2x − y + 3z = 9; check (1, 2, 3): 2 − 2 + 9 = 9 ✓. The three-point plane has intercepts 1, 1, 1: x + y + z = 1.
Common mistake: Using a point on the plane as the normal vector. The normal is perpendicular to the plane and read from the coefficients.
Practice:
  1. Form a plane equation through a point with a specified normal vector.
  2. Find the plane through (2, 1, 0), (3, 2, 1) and (1, 0, 3) using AB × AC.
  3. Find the plane through the intersection of x + y + z = 1 and 2x + 3y − z = 4 passing through (1, 1, 1).
Remember: The normal vector is perpendicular to every direction lying in the plane.
27

Angles and Distances in Space

Angles are found from direction vectors (lines) or normal vectors (planes). Between two planes, cos θ = |n₁·n₂|/(|n₁||n₂|); planes are parallel when normals are proportional and perpendicular when n₁·n₂ = 0. Between a line with direction b and a plane with normal n, sin θ = |b·n|/(|b||n|); the line is parallel to the plane if b·n = 0.

Distance formulas: from point (x₁, y₁, z₁) to plane ax + by + cz = d, distance = |ax₁ + by₁ + cz₁ − d|/√(a² + b² + c²); between parallel planes ax + by + cz = d₁ and = d₂: |d₁ − d₂|/√(a² + b² + c²); between a point and a line, and between skew lines, use the vector formulas of the previous topics.

Key formulas: Plane–plane: cos θ = |n₁·n₂|/(|n₁||n₂|)  |  Line–plane: sin θ = |b·n|/(|b||n|)  |  Point–plane: |ax₁ + by₁ + cz₁ − d|/√(a² + b² + c²)
Worked example: (a) Distance of P(1, 2, 3) from the plane 2x − y + 2z = 9. (b) Angle between x + y + z = 1 and x − y + z = 5. (c) Is the line with direction (1, 2, 2) parallel to the plane 2x + y − 2z = 7?
(a) |2 − 2 + 6 − 9|/√(4 + 1 + 4) = 3/3 = 1. (b) n₁ = (1, 1, 1), n₂ = (1, −1, 1); n₁·n₂ = 1; magnitudes √3 each; cos θ = 1/3, so θ = cos⁻¹(1/3). (c) b·n = 2 + 2 − 4 = 0, so the line is parallel to the plane.
Common mistake: Using cos instead of sin for the angle between a line and a plane. The angle with the plane is the complement of the angle with the normal.
Practice:
  1. Determine whether two planes are parallel by comparing their normal vectors.
  2. Find the distance between the parallel planes 2x + 2y − z = 3 and 2x + 2y − z = 9. (answer: 2)
  3. Find the angle between the line (x − 1)/1 = (y − 2)/2 = (z − 3)/2 and the plane x + y + z = 5.
Remember: Translate geometry into direction or normal vectors before applying angle formulas.

📊 Optimisation and Advanced Probability

28

Linear Programming and Feasible Regions

A linear programming problem (LPP) maximises or minimises a linear objective function Z = ax + by subject to linear constraints (inequalities) and non-negativity conditions x, y ≥ 0. The set of points satisfying all constraints is the feasible region, always convex. Solutions inside it are feasible; the best one is optimal.

The corner point theorem states that if the feasible region is bounded, the optimum of Z occurs at a corner (vertex). If the region is unbounded, a maximum or minimum may not exist; check by drawing the half-plane ax + by > M (or < m) and testing whether it has points common with the feasible region. If two corners give the same optimal value, every point on the segment between them is optimal (alternative optima). An empty feasible region means the problem is infeasible.

Procedure: (1) Write Z and the constraints. (2) Graph each inequality and shade the common region. (3) Find the corner points. (4) Evaluate Z at each. (5) Pick the maximum or minimum, and for unbounded regions verify with the half-plane test.
Worked example: Maximise Z = 3x + 2y subject to x + y ≤ 4, x + 3y ≤ 6, x, y ≥ 0.
Corner points: O(0, 0), A(4, 0), B(3, 1) (from x + y = 4 and x + 3y = 6), C(0, 2). Values of Z: 0, 12, 11, 4. The maximum is 12 at (4, 0).
Common mistake: Evaluating Z at points of the boundary that are not vertices, or forgetting the non-negativity constraints and shading the wrong side of a line (test the origin).
Practice:
  1. Graph two linear constraints and identify their common feasible region.
  2. Minimise Z = 2x + 3y subject to x + y ≥ 4, x ≥ 1, y ≥ 1 (answer: Z = 9 at (3, 1)).
  3. Why can the optimum of a bounded LPP not lie strictly in the interior of the feasible region?
Remember: Only points satisfying every constraint belong to the feasible region.
29

Conditional Probability and Independent Events

The probability of event A given that B has occurred is P(A|B) = P(A ∩ B)/P(B), provided P(B) > 0. Knowing B shrinks the sample space to B. Rearranging gives the multiplication rule P(A ∩ B) = P(B)P(A|B) = P(A)P(B|A). Properties include 0 ≤ P(A|B) ≤ 1, P(A′|B) = 1 − P(A|B), and P(E ∪ F | G) = P(E|G) + P(F|G) − P(E ∩ F | G).

Events A and B are independent if P(A ∩ B) = P(A)P(B), equivalently P(A|B) = P(A): the occurrence of one does not change the probability of the other. Three events are mutually independent only if every pair and the triple product satisfy the product rule. Independence is not the same as mutual exclusivity: mutually exclusive events with non-zero probabilities are always dependent, since one occurring makes the other impossible.

Key formulas: P(A|B) = P(A ∩ B)/P(B)  |  P(A ∩ B) = P(A)P(B|A)  |  independent: P(A ∩ B) = P(A)P(B)
Worked example: (a) Two dice are thrown. Find P(sum is 8 | first die shows 3). (b) Two cards are drawn without replacement. Find P(both aces). (c) A coin and a die are tossed: P(head and 4)?
(a) Given first die = 3, there are 6 outcomes; only (3, 5) gives sum 8, so the probability is 1/6. (b) (4/52)(3/51) = 1/221. (c) The events are independent: (1/2)(1/6) = 1/12.
Common mistake: Using P(A)P(B) for events that are not independent, as with drawing without replacement.
Practice:
  1. Use a two-way table to calculate a conditional probability.
  2. If P(A) = 0.6, P(B) = 0.5 and P(A ∩ B) = 0.3, are A and B independent? (answer: yes)
  3. A bag has 4 red and 6 black balls. Two are drawn without replacement. Find P(second red | first red). (answer: 3/9 = 1/3)
Remember: Independence and mutual exclusivity are different concepts.
30

Bayes’ Theorem and Random-Variable Distributions

If E₁, E₂, …, Eₙ are mutually exclusive and exhaustive events (a partition of the sample space) with P(Eᵢ) > 0, the theorem of total probability gives P(A) = ΣP(Eᵢ)P(A|Eᵢ). Bayes’ theorem reverses the conditioning: P(Eᵢ|A) = P(Eᵢ)P(A|Eᵢ)/ΣP(Eⱼ)P(A|Eⱼ). The P(Eᵢ) are prior probabilities and P(Eᵢ|A) the posterior probabilities after observing A.

A random variable X assigns a real number to each outcome. The probability distribution lists each value xᵢ with pᵢ = P(X = xᵢ), with Σpᵢ = 1. The mean (expectation) is E(X) = Σxᵢpᵢ and the variance Var(X) = E(X²) − [E(X)]², with standard deviation √Var. Bernoulli trials are independent trials with two outcomes (success p, failure q = 1 − p); the number of successes in n trials follows the binomial distribution B(n, p).

Key formulas: P(Eᵢ|A) = P(Eᵢ)P(A|Eᵢ)/ΣP(Eⱼ)P(A|Eⱼ)  |  E(X) = Σxᵢpᵢ  |  Var(X) = E(X²) − [E(X)]²
Binomial: P(X = r) = ⁿCr pr qn−r, mean = np, variance = npq
Worked example: (a) Machine A makes 60% of items with 2% defective; machine B makes 40% with 3% defective. A random item is defective; find P(from A). (b) Distribution of heads in two tosses and its mean and variance. (c) P(exactly 2 sixes in 4 throws of a die).
(a) P(A|D) = 0.6 × 0.02/(0.6 × 0.02 + 0.4 × 0.03) = 0.012/0.024 = 0.5. (b) X = 0, 1, 2 with probabilities 1/4, 1/2, 1/4; E(X) = 1; E(X²) = 0 + 1/2 + 1 = 1.5; Var = 1.5 − 1 = 0.5 (= npq = 2·½·½). (c) ⁴C₂(1/6)²(5/6)² = 6 × 25/1296 = 25/216 ≈ 0.116.
Common mistake: Confusing P(A|B) with P(B|A). Bayes’ theorem exists precisely because these differ.
Practice:
  1. Construct a simple probability distribution for the number of heads in two coin tosses.
  2. Find the mean of the number obtained on a single throw of a fair die. (answer: 3.5)
  3. A test detects a disease with 99% accuracy; 1% of people have it and the false-positive rate is 5%. Use Bayes’ theorem to find P(disease | positive). (answer: ≈ 0.167)
Remember: Probabilities in a discrete probability distribution must sum to 1.

📋 Formula Sheet and Exam Tips

Quick Revision Sheet

AreaKey results to memorise
Inverse trigonometrysin⁻¹x + cos⁻¹x = π/2; principal ranges; tan⁻¹x + tan⁻¹y = tan⁻¹[(x + y)/(1 − xy)]
MatricesA⁻¹ = adj A/|A|; (AB)′ = B′A′; (AB)⁻¹ = B⁻¹A⁻¹; |kA| = kⁿ|A|
DeterminantsArea of triangle = ½|det|; |AB| = |A||B|; row swap flips sign
Derivatives(uv)′ = u′v + uv′; (u/v)′ = (u′v − uv′)/v²; (xˣ)′ = xˣ(1 + ln x); inverse trig derivatives
Applications of derivativesTangent: y − y₀ = f′(x₀)(x − x₀); f′ > 0 increasing; f″ test for max/min
Integrals∫dx/(x² + a²) = (1/a)tan⁻¹(x/a); ∫eˣ[f + f′] = eˣf; ILATE for parts
Definite integrals∫abf(x)dx = ∫abf(a + b − x)dx; even/odd property
Differential equationsIF = e∫P dx; y = vx for homogeneous; separate variables
Vectorsa·b = |a||b|cos θ; |a × b| = |a||b|sin θ; [a b c] = a·(b × c)
3D geometrySkew-line distance |(a₂ − a₁)·(b₁ × b₂)|/|b₁ × b₂|; point–plane distance formula
ProbabilityP(A|B) = P(A ∩ B)/P(B); Bayes’ theorem; binomial mean np, variance npq

Exam strategy

  • Check domain and range first. Many errors in functions and inverse trigonometry come from ignoring principal values.
  • Show every step. Marks are given for method; write the formula, substitute, and state the result with units or sign where relevant.
  • Verify answers. Multiply a matrix by its inverse, differentiate an integral, or test a point in the feasible region.
  • Sketch before integrating. A rough graph prevents sign errors in area problems and shows which curve is above.
  • Watch limits and constants. Change limits during substitution, add + C for indefinite integrals, and apply initial conditions in differential equations.
  • Practise mixed problems daily. Rotate between algebra, calculus, vectors and probability so no chapter fades before the exam.

🎓 Class 12 Mathematics Complete

You have completed 30 detailed Class 12 Mathematics topics covering advanced functions, matrices, determinants, calculus, differential equations, vectors, 3D geometry, linear programming and advanced probability, each with formulas, worked examples, common mistakes and practice questions.

OnlyGyaani — Learn • Understand • Practise • Apply

Scroll to Top